Computing the nth derivative using the binomial theorem. Let n be an element of the set of  natural numbers
Let $F(x)=x^2(1+x)^n$ and write $F^n$ for the nth derivative of the function $F$. 
Compute $F^n$ by applying the Binomial Theorem to $(1+x)^n$.
I don't understand the step where I need to find the derivative of these terms:
= $x^2 + {n \choose 1}x^3 + {n \choose 2} x^4 + .... + {n \choose n-1} x^{n+1} + x^{n+2} $
Even though I do understand how to show that for the first 3 terms the nth derivative is 0, but for the last two I have no clue??
Really appreciate any hints!!
 A: The $n$-th derivative is $0$ for many more than just the first three terms: in fact, for all but the last three.
You have a typo, the last term is $x^{n+2}$.
The last three terms are $${n\choose{n-2}} x^n+{n\choose{n-1}} x^{n+1}+x^{n+2}$$
Can you work it out from here? Some factorials will appear when you take the $n$-th derivative, hence you might want to express the $n\choose k$ in terms of factorials as well, to see if something cancels out, and at the very least to get a nice-looking formula.
A: Since $\deg F=n+2$, we should get a quadratic. Use the generalized product rule$$\frac{d^n}{dx^n}g(x)h(x)=\sum_{k=0}^n\binom{n}{k}g^{(k)}h^{(n-k)}.$$For $g=x^2$, only $0\le k\le 2$ contributes, giving$$x^2\frac{d^n}{dx^n}(1+x)^n+2nx\frac{d^{n-1}}{dx^{n-1}}(1+x)^n+n(n-1)\frac{d^{n-2}}{dx^{n-2}}(1+x)^n.$$Since $\frac{d^{n-k}}{dx^{n-k}}(1+x)^n$ is $\frac{n!}{k!}(1+x)^k$, the above simplifies to$$n!x^2+n!2nx(1+x)+\frac{n!}{2}n(n-1)(1+x)^2\\=n!\left(\frac12n(n-1)+n(n+1)x+\frac12(n+1)(n+2)x^2\right).$$
A: Write it in decreasing powers of $x$ and use subscript $n$:
$$F_n(x)=x^2(1+x)^n=x^2(x+1)^n=\\
\color{red}{x^{n+2}}+\color{green}{nx^{n+1}}+\color{blue}{\frac{n(n-1)}{2}x^{n}}+\frac{n(n-1)(n-2)}{6}x^{n-1}+\cdots+nx^3+x^2$$
Note that for $f(x)=x^n$:
$$f^{(0)}(x)=f(x)=x^n\\
f^{(k)}(x)=\begin{cases}n(n-1)(n-2)\cdots(n-k+1)x^{n-k},k\le n\\ 0,k>n\end{cases}$$ 
Hence:
$$F_0(x)=x^2 \Rightarrow F_0^{(0)}(x)=F_0(x)=x^2;\\
F_n^{(n)}(x)=\color{red}{(n+2)(n+1)n\cdots3\cdot x^2}+\color{green}{n\cdot (n+1)n\cdots1\cdot x}+\\
\color{blue}{\frac{n(n-1)}{2}\cdot n(n-1)\cdots 1}=\\
\frac12(n+2)(n+1)\cdot n!\cdot x^2+n(n+1)\cdot n!\cdot x+\frac{n(n-1)}{2}\cdot n!=\\
n!\left(\frac12(n+2)(n+1)x^2+n(n+1)x+\frac{n(n-1)}{2}\right).$$
Verify:
$$F_1(x)=x^2(x+1)=x^3+x^2 \Rightarrow F_1^{(1)}(x)=1!(3x^2+2x)=3x^2+2x;\\
F_2(x)=x^2(x+1)^2=x^4+2x^3+x^2 \Rightarrow \\
F_2^{(2)}(x)=2!(6x^2+6x+1)=12x^2+12x+2.$$
